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Chemical Kinetics: JEE Main Chemistry Question with Solution The rate constant for a first order reaction is
20 min − 1 20\text{ min}^{-1} 20 min − 1 . The time required for the initial concentration of the reactant to reduce to its
1 32 \frac{1}{32} 32 1 level is
‾ 10 − 2 min \underline{\quad\quad\quad} 10^{-2}\text{ min} 1 0 − 2 min . (Nearest integer)
( Given : ln 10 = 2.303 (\text{Given }: \ln 10=2.303 ( Given : ln 10 = 2.303 log 2 = 0.3010 ) \log 2 = 0.3010) log 2 = 0.3010 ) No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.
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Choose one answer I’d rather see the answer directly Step-by-step solution View Correct answer
The time required is 17 × 10⁻² min, giving an integer value of 17. Option analysis
Why each option works or fails
Step Working
01 given Rate constant k = 20 min − 1 k = 20\text{ min}^{-1} k = 20 min − 1 [ A ] t [ A ] 0 = 1 32 \frac{[A]_t}{[A]_0} = \frac{1}{32} [ A ] 0 [ A ] t = 32 1 ln 10 = 2.303 \ln 10 = 2.303 ln 10 = 2.303 log 10 2 = 0.3010 \log_{10} 2 = 0.3010 log 10 2 = 0.3010
02 find Time t t t 10 − 2 min 10^{-2}\text{ min} 1 0 − 2 min
03 strategise For a first-order reaction, t = 1 k ln ( [ A ] 0 [ A ] t ) = 2.303 k log 10 ( 32 ) = 2.303 × 5 log 10 2 k t = \frac{1}{k}\ln\left(\frac{[A]_0}{[A]_t}\right) = \frac{2.303}{k}\log_{10}(32) = \frac{2.303 \times 5 \log_{10} 2}{k} t = k 1 ln ( [ A ] t [ A ] 0 ) = k 2.303 log 10 ( 32 ) = k 2.303 × 5 l o g 10 2
04 execute t = 2.303 × 5 × 0.3010 20 = 3.466 20 = 0.1733 min = 17.33 × 10 − 2 min ≈ 17 × 10 − 2 min t = \frac{2.303 \times 5 \times 0.3010}{20} = \frac{3.466}{20} = 0.1733\text{ min} = 17.33 \times 10^{-2}\text{ min} \approx 17 \times 10^{-2}\text{ min} t = 20 2.303 × 5 × 0.3010 = 20 3.466 = 0.1733 min = 17.33 × 1 0 − 2 min ≈ 17 × 1 0 − 2 min
✓ verify Rounding 17.33 17.33 17.33 17 17 17
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Question type Numerical
Exam relevance JEE Main · Chemistry
Concepts assessed Chemistry
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Can we directly use half-lives since 32 is a power of 2? Yes, reducing to 1 / 32 = ( 1 / 2 ) 5 1/32 = (1/2)^5 1/32 = ( 1/2 ) 5 t = 5 × t 1 / 2 = 5 × ln 2 k t = 5 \times t_{1/2} = 5 \times \frac{\ln 2}{k} t = 5 × t 1/2 = 5 × k l n 2
Answer The time required is 17 × 10⁻² min, giving an integer value of 17.
Why each option works or fails Step-by-step solution given: Rate constant k = 20 min − 1 k = 20\text{ min}^{-1} k = 20 min − 1 [ A ] t [ A ] 0 = 1 32 \frac{[A]_t}{[A]_0} = \frac{1}{32} [ A ] 0 [ A ] t = 32 1 ln 10 = 2.303 \ln 10 = 2.303 ln 10 = 2.303 log 10 2 = 0.3010 \log_{10} 2 = 0.3010 log 10 2 = 0.3010 find: Time t t t 10 − 2 min 10^{-2}\text{ min} 1 0 − 2 min strategise: For a first-order reaction, t = 1 k ln ( [ A ] 0 [ A ] t ) = 2.303 k log 10 ( 32 ) = 2.303 × 5 log 10 2 k t = \frac{1}{k}\ln\left(\frac{[A]_0}{[A]_t}\right) = \frac{2.303}{k}\log_{10}(32) = \frac{2.303 \times 5 \log_{10} 2}{k} t = k 1 ln ( [ A ] t [ A ] 0 ) = k 2.303 log 10 ( 32 ) = k 2.303 × 5 l o g 10 2 execute: t = 2.303 × 5 × 0.3010 20 = 3.466 20 = 0.1733 min = 17.33 × 10 − 2 min ≈ 17 × 10 − 2 min t = \frac{2.303 \times 5 \times 0.3010}{20} = \frac{3.466}{20} = 0.1733\text{ min} = 17.33 \times 10^{-2}\text{ min} \approx 17 \times 10^{-2}\text{ min} t = 20 2.303 × 5 × 0.3010 = 20 3.466 = 0.1733 min = 17.33 × 1 0 − 2 min ≈ 17 × 1 0 − 2 min verify: Rounding 17.33 17.33 17.33 17 17 17 Shortcut: When to use it: Whenever the remaining concentration fraction is an exact power of 1/2.
strategise: Since [ A ] t / [ A ] 0 = 1 / 32 = ( 1 / 2 ) 5 [A]_t / [A]_0 = 1/32 = (1/2)^5 [ A ] t / [ A ] 0 = 1/32 = ( 1/2 ) 5 n = 5 n = 5 n = 5 t = 5 × t 1 / 2 = 5 × ln 2 k t = 5 \times t_{1/2} = 5 \times \frac{\ln 2}{k} t = 5 × t 1/2 = 5 × k l n 2
execute: t = 5 × 2.303 × 0.3010 20 = 5 × 0.03466 = 0.1733 min = 17.33 × 10 − 2 min ≈ 17 × 10 − 2 min t = 5 \times \frac{2.303 \times 0.3010}{20} = 5 \times 0.03466 = 0.1733\text{ min} = 17.33 \times 10^{-2}\text{ min} \approx 17 \times 10^{-2}\text{ min} t = 5 × 20 2.303 × 0.3010 = 5 × 0.03466 = 0.1733 min = 17.33 × 1 0 − 2 min ≈ 17 × 1 0 − 2 min
verify: Integer value to report into the blank is 17.